Question: $\text F = \left[\begin{array}{r}5 \\ 0 \\ 4\end{array}\right]$ and $\text A = \left[\begin{array}{rr}-1 & 1\end{array}\right]$ Let $\text {H = FA}$. Find $\text H$. $ {H = }$
Explanation: The Strategy When multiplying matrices, we should find each entry of the resulting product matrix separately. To find entry $(i,j)$ of the resulting product matrix, we calculate the vector dot product of row $i$ of the first matrix and column $j$ of the second matrix. [I don't know what "vector dot product" is!] Finding $\text {H}_{1,1}$ $\text{H}_{1,1}$ is the dot product of the first row of $\text{F}$ and the first column of $\text{A}$. $ \text {H}=\left[\begin{array}{rr}{5} \\ 0 \\ 4\end{array}\right]\left[\begin{array}{rr} {-1} & 1\end{array}\right]$ Therefore, this is the appropriate calculation of $\text{H}_{1,1}$. $\begin{aligned}\text{H}_{1,1}&=(5)\cdot(-1)\\\\ &=-5 \end{aligned}$ The other entries of $\text{H}$ can be found similarly. Try it yourself for $\text{H}_{2,1}$ What is the appropriate calculation of ${H}_{2,1}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A ${H}_{2,1}$ does not exist. (Choice B) B $5 \cdot 1 = 5$ (Choice C) C $0 \cdot -1 = 0$ Check f Summary After calculating all the remaining entries of $\text{H}$, we get the following answer. $ \text {H}=\left[\begin{array}{rr}-5 & 5 \\ 0 & 0 \\ -4 & 4\end{array}\right] $